Does not exist cover of $\mathbb{R}^n$ by disjoint closed balls

For the Baire category proof.
Use the OP proof up to $$ C^{'}=\displaystyle\bigcup_{i=1}^{\infty} \partial B_{i} . $$ Then note $C^{'}$ is a complete metric space, written as a countable union of sets $\partial B_i$, which are closed (in $C'$) sets with empty interior (in $C'$).

To show no such cover via open balls exists is easy: any such cover $\mathcal{O}$ must involve at least 2 open balls (since the radii are finite). So pick $O\in\mathcal{O}$, and let

$A=O$,
$B=\bigcup (\mathcal{O}\setminus\{O\})$.

Then $A$ is open by assumption, while $B$ is open as a union of open sets; and $A\sqcup B=\mathbb{R}^n$. But this contradicts the connectivity of $\mathbb{R}^n$.

Here's one way to do it. Suppose towards contradiction that $\mathcal{U}$ is a set of countably many disjoint closed balls with finite positive radius, which covers $\mathbb{R}^n$ (you've justified in your question why we may assume the cover is countable). Say that a closed ball $B$ is shattered if it is not contained in a single element of $\mathcal{U}$.

Exercise: there is a shattered closed ball.

Now the crucial point is:

If $B$ is a shattered closed ball, and $U\in\mathcal{U}$, then there is a closed ball $C$ of aribtrarily small positive radius contained in $B\setminus U$ which is also shattered.

Why? Well, clearly there are $x, y\in B\setminus U$ which lie in different elements $V_x, V_y$ of $\mathcal{U}$ (why?). This means that $\partial V_x\cap (B\setminus U)$ is nonempty (why?); let $z$ be an element of this boundary. Then an appropriately small-radius closed ball around $z$ is shattered.

OK, so what? Well, now we can diagonalize against $\mathcal{U}$! Let $\mathcal{U}=\{U_i: i\in\mathbb{N}\}$, and define a sequence of shattered closed balls of finite positive radius $C_i$ such that